370 Higher Engineering Mathematics
(b) Rearranging
(1 − t )
2 dt gives:
(1 − 2t + t
2
) dt = t −
2t 1+1
1 + 1
+
t 2+1
2 + 1
+ c
= t −
2t 2
2
+
t 3
3
+ c
= t − t
2
+
1
3
t
3
+ c
This problem shows that functions often have to be
rearranged into the standard form of
ax
n dx before
it is possible to integrate them.
Problem 4. Determine
3
x 2 dx.
3
x 2 dx =
3x
−2 dx. Using the standard integral,
ax
n dx when a = 3 and n =−2 gives:
3x
−2 dx =
3x
−2+1
−2 + 1
+ c =
3x
−1
−1
+ c
= −3x
−1
+ c =
−3
x
+ c
Problem 5. Determine
3
√
x dx.
For fractional powers it is necessary to appreciate
n
√
a m = a
m
n
3
√
x dx =
3x
1
2 dx =
3x
1
2 +1
1
2
+ 1
+ c
=
3x
3
2
3
2
+ c = 2x
3
2 + c = 2
x 3 + c
Problem 6. Determine
−5
9
4
√
t 3
dt .
−5
9
4
√
t 3
dt =
−5
9t
3
4
dt =
−
5
9
t
− 3
4 dt
=
−
5
9
t
−
3
4
+1
−
3
4
+ 1
+ c
=
−
5
9
t
1
4
1
4
+ c =
−
5
9
4
1
t
1
4 + c
= −
20
9
4
√
t + c
Problem 7. Determine
(1 + θ) 2
√ θ
dθ.
(1 + θ) 2
√
θ
dθ =
(1 + 2θ + θ 2 )
√
θ
dθ
=
1
θ
1
2
+
2θ
θ
1
2
+
θ 2
θ
1
2
dθ
=
θ
−1
2 + 2θ
1−
1
2
+ θ
2−
1
2
dθ
=
θ
−1
2 + 2θ
1
2 + θ
3
2
dθ
=
θ
−1
2
+1
−
1
2 + 1
+
2θ
1
2
+1
1
2 + 1
+
θ
3
2
+1
3
2 + 1
+ c
=
θ
1
2
1
2
+
2θ
3
2
3
2
+
θ
5
2
5
2
+ c
= 2θ
1
2 +
4
3
θ
3
2 +
2
5
θ
5
2 + c
= 2
√
θ +
4
3
θ
3
+
2
5
θ
5
+ c
Problem 8. Determine
(a)
4 cos3x dx (b)
5 sin2θ dθ.
(a) From Table 37.1(ii),
4 cos3x dx = (4)
1
3
sin 3x + c
=
4
3
sin 3x + c
(b) From Table 37.1(iii),
5 sin2θ dθ = (5)
−
1
2
cos 2θ + c
= −
5
2
cos 2θ + c